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Library / Dimensional analysis and similarity / DIM-20

Dimensional analysis and similarity

Pump or fan shaft power at a new speed, diameter and density

Affinity law for shaft power: for geometrically similar machines at corresponding operating points and comparable efficiency, power is proportional to density, the cube of the speed and the fifth power of the diameter.

Ref
DIM-20
Rev
v1
Field
Dimensional analysis and similarity
Result
P2 [kW]

With s = n2/n1, d = D2/D1 and z = ρ2/ρ1, the shaft power at the corresponding operating point scales as P2 = P1 z s³ d⁵. It assumes comparable efficiency at the two points. These affinity laws compare corresponding operating points of geometrically similar rotodynamic machines. They assume low-compressibility fan flow or non-cavitating pump flow. The diameter powers describe a fully scaled machine; trimming an impeller in an existing casing changes the geometry ratios and needs a trim curve or manufacturer data.

Affinity law for shaft power: for geometrically similar machines at corresponding operating points and comparable efficiency, power is proportional to density, the cube of the speed and the fifth power of the diameter.

Calculator

Solve for
kW
kW
0.015,000
rpm
5010,000
rpm
5010,000
mm
203,000
mm
203,000
kg/m³
0.12,000
kg/m³
0.12,000
  • Speed ratio, s1.2
  • Diameter ratio, d1
  • Power multiplier, z s³ d⁵1.728
  • Power if the new speed is reduced by 20 %5.308 kW

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Shaft power at point 2 (kW) current value 10.37 kW
D1Impeller diameter at point 1Key driver
1.04e-47,870,000
D2Impeller diameter at point 2
1.37e-51,040,000
ρ1Fluid density at point 1
5.18104,000
n1Speed at point 1
0.010482,900
P1Shaft power at point 1
0.01738,640
n2Speed at point 2
7.5e-46,000
ρ2Fluid density at point 2
0.0010420.7
  • Raising the speed by a factor of 1.2 (with the same diameter and fluid) multiplies the power by 1.728, from 6 to 10.37 kW: a 20 % increase in speed costs 73 % more power.
  • The cube law cuts both ways: slowing a pump or fan by 20 % cuts its power to about half, which is why variable-speed drives save so much energy in part-load duty.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kW)Local stepChange in result
D1 Impeller diameter at point 120 to 3,000 mm1.037e-47,873,0007,873,000+10%−37.9 %
D2 Impeller diameter at point 220 to 3,000 mm1.365e-51,037,0001,037,000+10%+61.1 %
ρ1 Fluid density at point 10.1 to 2,000 kg/m³5.184103,700103,700+10%−9.09 %
n1 Speed at point 150 to 10,000 rpm0.0103782,94082,940+10%−24.9 %
P1 Shaft power at point 10.01 to 5,000 kW0.017288,6408,640+10%+10 %
n2 Speed at point 250 to 10,000 rpm7.5e-46,0006,000+10%+33.1 %
ρ2 Fluid density at point 20.1 to 2,000 kg/m³0.00103720.7420.73+10%+10 %

Sweep of Speed at point 2

Speed at point 2 (rpm)Shaft power at point 2 (kW)
507.5e-4
1,29412.99
2,53898.03
3,781324.4
5,025761.3
6,2691,478
7,5132,544
8,7564,028
10,0006,000

Sweep one variable

Shaft power at point 2 (kW)Hover, touch, or use the arrow keys on the chart
0100020003000400050006000 200040006000800010k Speed at point 2 (rpm) now 10.37

Variables and how each one contributes

P2

Shaft power at point 2

Result · kW

What the equation gives you. Everything else feeds it.

P1

Shaft power at point 1

kW · typical 0.01 to 5,000

The shaft power at the original operating point, in kW. Raises the new power in direct proportion. Raising it by 10 % (to 6.6) moves the result by +10 %, from 10.37 to 11.4 kW.

n1

Speed at point 1

rpm · typical 50 to 10,000

The original rotational speed, in rpm. Lowers the new power with the cube of the speed ratio. Raising it by 10 % (to 1,100) moves the result by −24.9 %, from 10.37 to 7.79 kW.

n2

Speed at point 2

rpm · typical 50 to 10,000

The new rotational speed, in rpm. Raises the new power with the cube of the speed, so speed is a strong lever. Raising it by 10 % (to 1,320) moves the result by +33.1 %, from 10.37 to 13.8 kW.

D1

Impeller diameter at point 1

mm · typical 20 to 3,000

The original impeller diameter, in mm. Lowers the new power with the fifth power of the diameter ratio. Raising it by 10 % (to 330) moves the result by −37.9 %, from 10.37 to 6.438 kW.

D2

Impeller diameter at point 2

mm · typical 20 to 3,000

The new impeller diameter, in mm. Raises the new power with the fifth power of the diameter, so diameter is the strongest lever. Raising it by 10 % (to 330) moves the result by +61.1 %, from 10.37 to 16.7 kW.

ρ1

Fluid density at point 1

kg/m³ · typical 0.1 to 2,000

The fluid density at the original point, in kg/m³. Lowers the new power in inverse proportion. Raising it by 10 % (to 1,100) moves the result by −9.09 %, from 10.37 to 9.425 kW.

ρ2

Fluid density at point 2

kg/m³ · typical 0.1 to 2,000

The fluid density at the new point, in kg/m³. Raises the new power in direct proportion. Raising it by 10 % (to 1,100) moves the result by +10 %, from 10.37 to 11.4 kW.

Worked example with the values above

  1. Ratios: s = 1.2, d = 1 and z = 1.
  2. Power multiplier: z s³ d⁵ = 1 × 1.2³ × 1⁵ = 1.728.
  3. P2 = 6 × 1.728 = 10.368 kW.

Assumptions and limits

  • Geometric similarity and corresponding operating points, with comparable efficiency at the two points.
  • Low-compressibility fan flow or non-cavitating pump flow.

Common mistakes

  • Sizing the motor for the scaled power without allowing for a change in efficiency or a margin.
  • Applying the fifth-power diameter law to a trimmed impeller.
  • Using the electrical input power at one point and the shaft power at the other.

Statistics and uncertainty

The result is a product of powers, so to first order its relative error is the root sum of squares of each input’s relative error multiplied by its exponent (P1 × 1, ρ2 × 1, n2 × 3, n1 × −3, D2 × 5, D1 × −5). The diameter (fifth power) and speed (cube) errors dominate, and efficiency changes between the two points add a few per cent on top of that.

Where it is used

  • Estimating the power a pump or fan needs after a speed, size or density change.
  • Estimating the energy saved by running a pump or fan more slowly.

Maths you need

  • Shaft power: power delivered to the impeller by the shaft
  • Efficiency: useful power divided by shaft power

Related formulas

Practise this formulaOpen the dimension solver